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Nigerian Spleen Length. O.Ehimwenma and M.Tagho, researchers in Nigeria, were interested in how characteristics of the spleen of residents in their tropical environment compare to those found elsewhere in te world. They published their findings in the article "Determination of Normal Dimensions of the Spleen by Ultrasound in an Endemic Tropical Environment" (Nige4ian Medical Journal, Vol.52, No. 3,pp. 198-203). The researchers randomly samples 91males and 109females in Nigeria. The mean and standard deviation of the spleen lengths for the males were 11.1cm and 0.9cm respectively, and those for the females were 10.1cm and 0.7cm, respectively. At the 1%significance level, do the data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians?

Short Answer

Expert verified

The data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians at1%significance level.

Step by step solution

01

Step 1. Given information 

Significance level=1%.

02

Step 2. Definition for Null and Alternative hypothesis:

Null Hypothesis:

H0:μ1=μ2

That is, the data does not provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians.

Alternative Hypothesis:

H0:μ1μ2

That is, the data provide sufficient evidence to conclude that a difference exists in mean spleen lengths of male and female Nigerians.

03

Step 3. The Given significance level is, α=0.01.

Now compute the value of the test statistic by using MINITAB.

MINITAB Procedure:

  • Choose Stat > Basic Statistics > 2-Sample t.
  • Choose Summarized data.
  • In first, enter Sample size as 91, Mean as 11.1, Standard deviation as 0.9.
  • In second, enter Sample size as 109, Mean as 10.1, Standard deviation as 0.7.
  • Select Assume equal variances.
  • Choose options.
  • In confidence level enter 99.
  • In Alternative select not equal.
  • Click OK in all the dialogue boxes.
04

Step 4. MINITAB output:

Two-Sample T-Test and CI

From the MINITAB output, the value of test statistic is8.83.

05

Step 5. Find the P-value:

From the MINITAB output, the P-value is, 0.0000.

If Pα, then reject the null hypothesis. Here, the P-value is 0.000which is less than the level of significance. That is, P=0.000<α=0.01.

Therefore, the null hypothesis is rejected at 1%level. Thus, it can be conclude that the test results are statistically significant at1%significance level.

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Most popular questions from this chapter

In each of Exercises 10.35-10.38, we have provided summary statistics for independent simple random samples from two populations. Preliminary data analyses indicate that the variable under consideration is normally distributed on each population. Decide, in each case, whether use of the pooled t-lest and pooled t-interval procedure is reasonable. Explain your answer.
10.35 x¯1=468.3,s1=38.2,n1=6

x2=394.6,s2=84.7,n2=14


In each of Exercises 10.75-10.80, we have provided summary statistics for independent simple random samples from non populations. In each case, use the non pooled t-test and the non pooled t-interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

x¯1=15,s1=2,n1=15,x¯2=12,s2=5andn2=15.

a. Two-tailed test, α=0.05

b. 95%confidence interval.

Define the phrase independent samples.

Right-Tailed Hypothesis Tests and CIs. If the assumptions for a pooled t-interval are satisfied, the formula for a (1-α)-level lower confidence bound for the difference, μ1-μ2, between two population means is

x¯1-x^2-ta·Sp1/n1+1/n2

For a right-tailed hypothesis test at the significance level α,

the null hypothesis H0:μ1=μ2will be rejected in favor of the alternative hypothesis Ha:μ1>μ2if and only if the (1-α)-level lower confidence bound for μ1-μ2is greater than or equal to 0. In each case, illustrate the preceding relationship by obtaining the appropriate lower confidence bound and comparing the result to the conclusion of the hypothesis test in the specified exercise.

a. Exercise 10.47

b. Exercise 10.50

Provide an example (different from the ones considered in this section) of a procedure based on a paired sample being more appropriate than one based on independent samples.

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